YES

We show the termination of the TRS R:

  norm(nil()) -> |0|()
  norm(g(x,y)) -> s(norm(x))
  f(x,nil()) -> g(nil(),x)
  f(x,g(y,z)) -> g(f(x,y),z)
  rem(nil(),y) -> nil()
  rem(g(x,y),|0|()) -> g(x,y)
  rem(g(x,y),s(z)) -> rem(x,z)

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: norm#(g(x,y)) -> norm#(x)
p2: f#(x,g(y,z)) -> f#(x,y)
p3: rem#(g(x,y),s(z)) -> rem#(x,z)

and R consists of:

r1: norm(nil()) -> |0|()
r2: norm(g(x,y)) -> s(norm(x))
r3: f(x,nil()) -> g(nil(),x)
r4: f(x,g(y,z)) -> g(f(x,y),z)
r5: rem(nil(),y) -> nil()
r6: rem(g(x,y),|0|()) -> g(x,y)
r7: rem(g(x,y),s(z)) -> rem(x,z)

The estimated dependency graph contains the following SCCs:

  {p1}
  {p2}
  {p3}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: norm#(g(x,y)) -> norm#(x)

and R consists of:

r1: norm(nil()) -> |0|()
r2: norm(g(x,y)) -> s(norm(x))
r3: f(x,nil()) -> g(nil(),x)
r4: f(x,g(y,z)) -> g(f(x,y),z)
r5: rem(nil(),y) -> nil()
r6: rem(g(x,y),|0|()) -> g(x,y)
r7: rem(g(x,y),s(z)) -> rem(x,z)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      g > norm#
    
    argument filter:
  
      pi(norm#) = [1]
      pi(g) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(x,g(y,z)) -> f#(x,y)

and R consists of:

r1: norm(nil()) -> |0|()
r2: norm(g(x,y)) -> s(norm(x))
r3: f(x,nil()) -> g(nil(),x)
r4: f(x,g(y,z)) -> g(f(x,y),z)
r5: rem(nil(),y) -> nil()
r6: rem(g(x,y),|0|()) -> g(x,y)
r7: rem(g(x,y),s(z)) -> rem(x,z)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      g > f#
    
    argument filter:
  
      pi(f#) = 2
      pi(g) = [1, 2]

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: rem#(g(x,y),s(z)) -> rem#(x,z)

and R consists of:

r1: norm(nil()) -> |0|()
r2: norm(g(x,y)) -> s(norm(x))
r3: f(x,nil()) -> g(nil(),x)
r4: f(x,g(y,z)) -> g(f(x,y),z)
r5: rem(nil(),y) -> nil()
r6: rem(g(x,y),|0|()) -> g(x,y)
r7: rem(g(x,y),s(z)) -> rem(x,z)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      rem# > s > g
    
    argument filter:
  
      pi(rem#) = 2
      pi(g) = []
      pi(s) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.